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                <span class="article-type type-1 float-left">原</span>                <h1 class="title-article">mysql查询最接近的记录</h1>
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                                                                                                                                            <span class="time">2017年09月13日 09:48:56</span>
                    <a class="follow-nickName" href="https://me.csdn.net/mxdzchallpp" target="_blank">CyborgLin</a>
                    <span class="read-count">阅读数 1504</span>
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                        版权声明：本文为博主原创文章，未经博主允许不得转载。                        <a class="copy-right-url" href=" https://blog.csdn.net/mxdzchallpp/article/details/77961905"> https://blog.csdn.net/mxdzchallpp/article/details/77961905</a>
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<p>查询场景：现在的需求是查询年龄最接近20岁的用户，获取前5个</p>
<p>我现在的数据库记录用户年龄的字段记录格式是"1995-05-20"，字段名称birthday</p>
<p>解决思路：</p>
<p>1.首先查询时转换成用户年龄</p>
<p>日期格式转年龄的方法：</p>
<p>（1）当前年份 - 日期格式中的年份</p>
<p></p><pre><code class="language-sql">date_format(now(), '%Y') - from_unixtime(unix_timestamp(birthday), '%Y')</code></pre>（2）当前年份 - 日期格式中的年份（获取方法不同）
<p></p><pre><code class="language-sql">date_format(now(), '%Y') - year(birthday)</code></pre>2.通过order by asb() 排序获取最接近的数据
<p>完整的语句：</p>
<p></p><pre><code class="language-sql">SELECT
	(date_format(now(), '%Y') - year(birthday)) as age
FROM
	user
WHERE
	is_anchor = 1
ORDER BY abs(20 - age)
limit 0,5</code></pre><br /><br /><br /><br />                                    </div>
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